The type of map
is map :: (a -> b) -> [a] -> [b]
. Hence the type of map foo
is obtained from [a] -> [b]
by substituting a
and b
with what can be derived from foo
's type. If, for example, foo :: t -> t
, you substitute a = t, b = t
and obtain [t] -> [t]
. If foo :: [t] -> Int
, you obtain [[t]] -> [Int]
.
In your case, the type of foo
(which is map
) is (x -> y) -> [x] -> [y]
. You have to unify that type with a -> b
to find out what a
and b
have to be substituted with. [Note that the function arrow is right-associative, x -> y -> z = x -> (y -> z)
.]
To find the type of
\x -> x >>= (\y -> y)
use the known type of (>>=) :: Monad m => m a -> (a -> m b) -> m b
. Ignore the constraint (Monad m =>
) for now.
As the first argument of (>>=)
, x
must have a type m a
for as yet unknown m
and a
. The second argument of (>>=)
is here the identity,
(\y -> y) :: t -> t
so you must unify t -> t
with a -> m b
. That gives you some information about a
, namely a = m b
.
That gives x :: m (m b)
, and (\x -> x >>= (\y -> y)) :: type_of_x -> type_of_rhs
.
Finally remember the temporarily forgotten constraint Monad m =>
.